Question Number 152147 by john_santu last updated on 26/Aug/21
Answered by Ar Brandon last updated on 26/Aug/21
![I_n =∫_(−1) ^1 ((√(x^2 +(1/n)))−∣x∣)dx =2∫_0 ^1 ((√(x^2 +(1/n)))−∣x∣)dx =2∫_0 ^(argsh((√n))) ((1/n)(√(sinh^2 ϑ+1)))coshϑdϑ−1 =(1/n)∫_0 ^(argsh((√n))) (cosh2ϑ+1)dϑ−1 =(1/n)[((sinh(2ϑ))/2)+ϑ]_0 ^(argsh((√n))) −1 =(1/n)(((sh(2argsh((√n))))/2)+argsh((√n)))−1=0](https://www.tinkutara.com/question/Q152149.png)
$${I}_{{n}} =\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{n}}}−\mid{x}\mid\right){dx} \\ $$$$\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{n}}}−\mid{x}\mid\right){dx} \\ $$$$\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{argsh}\left(\sqrt{{n}}\right)} \left(\frac{\mathrm{1}}{{n}}\sqrt{\mathrm{sinh}^{\mathrm{2}} \vartheta+\mathrm{1}}\right)\mathrm{cosh}\vartheta{d}\vartheta−\mathrm{1} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{argsh}\left(\sqrt{{n}}\right)} \left(\mathrm{cosh2}\vartheta+\mathrm{1}\right){d}\vartheta−\mathrm{1} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{n}}\left[\frac{\mathrm{sinh}\left(\mathrm{2}\vartheta\right)}{\mathrm{2}}+\vartheta\right]_{\mathrm{0}} ^{\mathrm{argsh}\left(\sqrt{{n}}\right)} −\mathrm{1} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{sh}\left(\mathrm{2argsh}\left(\sqrt{{n}}\right)\right)}{\mathrm{2}}+\mathrm{argsh}\left(\sqrt{{n}}\right)\right)−\mathrm{1}=\mathrm{0} \\ $$